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Sagot :
To calculate the final enthalpy of the overall chemical equation [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex], we will use Hess's Law, which states that the enthalpy change of the overall reaction is the sum of the enthalpies of the individual steps.
Given intermediate reactions and their enthalpies are:
[tex]\[ \begin{array}{ll} C(s) + O_2(g) \rightarrow CO_2(g) & \Delta H_1 = -393.5 \text{kJ} \\ 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2 = -566.0 \text{kJ} \\ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) & \Delta H_3 = 483.6 \text{kJ} \end{array} \][/tex]
1. Step 1: Reverse the second equation to get the formation of carbon monoxide from carbon dioxide:
[tex]\[ 2 CO_2(g) \rightarrow 2 CO(g) + O_2(g) \quad \Delta H_2' = 566.0 \text{kJ} \][/tex]
Here, the enthalpy changes sign because the reaction direction is reversed.
2. Step 2: Reverse the third equation to get the formation of water from hydrogen and oxygen:
[tex]\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \quad \Delta H_3' = -483.6 \text{kJ} \][/tex]
Again, the enthalpy changes sign because the reaction direction is reversed.
3. Step 3: Combine the enthalpy changes. However, since we need only 1 mole of [tex]\( CO \)[/tex] and [tex]\( H_2O \)[/tex] each:
[tex]\[ \frac{1}{2} \left( 2 CO_2(g) \rightarrow 2 CO(g) + O_2(g) \right) = CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad \Delta H_2'' = \frac{566.0}{2} = 283.0 \text{kJ} \][/tex]
[tex]\[ \frac{1}{2} \left( 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \right) = H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad \Delta H_3'' = \frac{-483.6}{2} = -241.8 \text{kJ} \][/tex]
4. Step 4: Add the enthalpies of these adjustments to [tex]\( \Delta H_1 \)[/tex]:
[tex]\[ \Delta H_{final} = \Delta H_1 + \Delta H_2'' + \Delta H_3'' \][/tex]
[tex]\[ \Delta H_{final} = -393.5 + 283.0 - 241.8 = -352.3 \text{kJ} \][/tex]
Thus, the enthalpy change for the overall chemical equation [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] is [tex]\(\boxed{-352.3 \text{kJ}}\)[/tex].
To summarize:
1. Reverse the second equation and change the sign of the enthalpy.
2. Reverse and half the third equation and change the sign of its enthalpy.
3. Sum the enthalpies through the given process steps.
This calculation leads us to the final enthalpy change of [tex]\(-352.3\)[/tex] kJ.
Given intermediate reactions and their enthalpies are:
[tex]\[ \begin{array}{ll} C(s) + O_2(g) \rightarrow CO_2(g) & \Delta H_1 = -393.5 \text{kJ} \\ 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2 = -566.0 \text{kJ} \\ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) & \Delta H_3 = 483.6 \text{kJ} \end{array} \][/tex]
1. Step 1: Reverse the second equation to get the formation of carbon monoxide from carbon dioxide:
[tex]\[ 2 CO_2(g) \rightarrow 2 CO(g) + O_2(g) \quad \Delta H_2' = 566.0 \text{kJ} \][/tex]
Here, the enthalpy changes sign because the reaction direction is reversed.
2. Step 2: Reverse the third equation to get the formation of water from hydrogen and oxygen:
[tex]\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \quad \Delta H_3' = -483.6 \text{kJ} \][/tex]
Again, the enthalpy changes sign because the reaction direction is reversed.
3. Step 3: Combine the enthalpy changes. However, since we need only 1 mole of [tex]\( CO \)[/tex] and [tex]\( H_2O \)[/tex] each:
[tex]\[ \frac{1}{2} \left( 2 CO_2(g) \rightarrow 2 CO(g) + O_2(g) \right) = CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad \Delta H_2'' = \frac{566.0}{2} = 283.0 \text{kJ} \][/tex]
[tex]\[ \frac{1}{2} \left( 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \right) = H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad \Delta H_3'' = \frac{-483.6}{2} = -241.8 \text{kJ} \][/tex]
4. Step 4: Add the enthalpies of these adjustments to [tex]\( \Delta H_1 \)[/tex]:
[tex]\[ \Delta H_{final} = \Delta H_1 + \Delta H_2'' + \Delta H_3'' \][/tex]
[tex]\[ \Delta H_{final} = -393.5 + 283.0 - 241.8 = -352.3 \text{kJ} \][/tex]
Thus, the enthalpy change for the overall chemical equation [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] is [tex]\(\boxed{-352.3 \text{kJ}}\)[/tex].
To summarize:
1. Reverse the second equation and change the sign of the enthalpy.
2. Reverse and half the third equation and change the sign of its enthalpy.
3. Sum the enthalpies through the given process steps.
This calculation leads us to the final enthalpy change of [tex]\(-352.3\)[/tex] kJ.
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